Domain and range? (1 Viewer)

[4028023]

So i understand how you solve the basics of domain for instance:

If the equation is normal eg ax^2+bx+c,then it would be considered all real x
If the equation is is under a square root, then let the equations be > or = 0
If the equation has a denominator, then the equation cannot equal 0

And there are some different strategies applied to the above based on different circumstances.

But how do you solve for range?Im completely confused because do we graph the equation for range?Is there a simple easy method?

:wave:
Fast response would be nice..thanks in advance

 

[Drongoski]

So i understand how you solve the basics of domain for instance:

If the equation is normal eg ax^2+bx+c,then it would be considered all real x
If the equation is is under a square root, then let the equations be > or = 0
If the equation has a denominator, then the equation cannot equal 0

And there are some different strategies applied to the above based on different circumstances.

But how do you solve for range?Im completely confused because do we graph the equation for range?Is there a simple easy method?

:wave:
Fast response would be nice..thanks in advance

For simple case of f(x) = x² - 3
the domain is the set of all real numbers and the range is all real values >= -3 since the smallest value f(x) may have is -3. The range of a function for a given or assumed domain is simply the set of all values it can take.

Why not post a few sample questions ?

 

[4028023]

y=1/x
y=x^2
y=3x^2+4x+5


how would we find the range of these?

 

[Timothy.Siu]

y=1/x
y=x^2
y=3x^2+4x+5


how would we find the range of these?

umm u see if theres any restrictions, or for these ones u can usually just picture them in your head to help, or draw them.
for y=1/x y can be all real numbers except 0
y=x^2 y>=0 since x^2 is always positive
y=3x^2+4x+5 find the turning point.

 

[Drongoski]

y=1/x
y=x^2
y=3x^2+4x+5


how would we find the range of these?

Further to Timothy's:

1) y = x^2 is an upright parabola, whose lowest value is 0

2) y = 3x^2+4x+5 is another upright parabola and has a minimum 11/3. This minimum value can be found by Tim's approach, by finding the value of the vertex (turning point) at x= "-b/2a" = -4/(2x3) = -2/3 or by completion of the square (quite messy for this case)